primes and unique factorization

**empressA** · January 20th 2010, 12:37 PM

Prove or disprove each of the following statements:

a. if p is prime and p divides (a^2 + b^2) and p divides (c^2 + d^2), then p divides (a^2 - c^2).

b. if p is prime and p divides (a^2 + b^2) and p divides (c^2 + d^2), then p divides (a^2 + c^2).

c. If p is prime and p divides a and p divides (a^2 + b^2), then p divides b.

help plz..dont know where to start..

**Drexel28** · January 20th 2010, 12:40 PM

Originally Posted by empressA

Prove or disprove each of the following statements:

a. if p is prime and p divides (a^2 + b^2) and p divides (c^2 + d^2), then p divides (a^2 - c^2).

b. if p is prime and p divides (a^2 + b^2) and p divides (c^2 + d^2), then p divides (a^2 + c^2).

c. If p is prime and p divides a and p divides (a^2 + b^2), then p divides b.

help plz..dont know where to start..

What do you know about primes? Let's start with that.

**empressA** · January 20th 2010, 12:46 PM

I know that p is a prime only if its divisors are positive neg. 1 and pos.neg. p..

i also know that if p is a prime and q is a prime then p divides q

and if p is a prime number that divides bc then it p divides b and p divides c..

**Drexel28** · January 20th 2010, 12:49 PM

Originally Posted by empressA

I know that p is a prime only if its divisors are positive neg. 1 and pos.neg. p..

i also know that if p is a prime and q is a prime then p divides q

and if p is a prime number that divides bc then it p divides b and p divides c..

That second statement is scarily wrong. $2,3$ .

**Dinkydoe** · January 20th 2010, 12:50 PM

This question should rather be posted in the number-theory section.

(a) is false: let p= 5, a= 1, b= 2, c= 3, d = 1

Have you tried a few examples?

(b) Find a counter-example for (b).

(c) Is correct. If $a\equiv 0$ mod p then $a^2+b^2\equiv b^2$ mod p. Thus $(a^2+b^2)\equiv 0$ mod p $\Rightarrow b \equiv 0$ mod p.

**empressA** · January 20th 2010, 12:52 PM

thanks dinky doe..

drexel 128..this is the notes my lecturer gave..lol..at least that second and third line

**empressA** · January 20th 2010, 01:01 PM

Originally Posted by Dinkydoe

(c) Is correct. If $a\equiv 0$ mod p then $a^2+b^2\equiv b^2$ mod p. Thus $(a^2+b^2)\equiv 0$ mod p $\Rightarrow b \equiv 0$ mod p.

we haven't reached congruence modular arithmetic

**Dinkydoe** · January 20th 2010, 01:01 PM

Empress. I can hardly believe that's what your lecturer gave.

The very definition of prime-numbers is that they can't be divided by numbers other then themself and 1.

Thus if p,q are primenumbers and $p|q \Rightarrow p=q$ . I think that's what your lecturer meant.

**empressA** · January 20th 2010, 01:04 PM

yea..thats what i have..but i took it as p / q..because he didn't differentiate between the sign between the p and the q..but i c now

**Dinkydoe** · January 20th 2010, 01:08 PM

we haven't reached congruence modular arithmetic

Then try to prove the following:

If $p|a$ , this means $a = np$ for some $n\geq 0$ . So $a^2 = n^2p^2$ .
And if $p|(a^2+b^2)\Rightarrow a^2+b^2 = pm$

Thus if $a^2+b^2 = n^2p^2+ b^2 = pm$ . What does this say about $b^2$ ?

**empressA** · January 20th 2010, 01:18 PM

Thus if $a^2+b^2 = n^2p^2+ b^2 = pm$ . What does this say about $b^2$ ?[/quote]

that b is zero

**Dinkydoe** · January 20th 2010, 01:22 PM

No:

Observe that from the above equation follows: $b^2 = p(m-n^2p)\Rightarrow p|b^2$

Now use the definition of prime-numbers.

**empressA** · January 20th 2010, 01:24 PM

that would imply that p divides a^2 + b^2

**Dinkydoe** · January 20th 2010, 01:26 PM

that would imply that p divides a^2 + b^2

No, that's what we assumed in the first place. We also assumed that p divides a.

In the last post I showed you that from these assumptions follows that p divides $b^2$ .

The definition of prime-numbers gives you that p divides b.

**empressA** · January 20th 2010, 01:33 PM

ok..i understand where you're coming from now. so since p / b^2 it also divides b..

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